integration by parts, DI method, VERY EASY

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Recommends Dinno Koluh · 2019-11 · Mathematics Stack Exchange · on “Calculating an Integral by Integration by parts”

I would suggest you to watch this video ( [this video] ) where the DI method on integration by parts is explained. It is a faster and more elegant method for dealing with integration by parts but it is based on the same idea. I think that you will find it useful for future work.

Recommends Joe · 2021-02 · Mathematics Stack Exchange · on “Use integration by parts twice $\int_{0}^{\frac{\pi}{8}}x^2\cos(4x)dx$”

@Emily: BlackPenRedPen has a video [this video] on it. You don't need it for A-level, but it does speed things up,

Recommends Joshua Tilley · 2023-10 · Mathematics Stack Exchange · on “Question regarding "integration by parts"”

Assume that we can differentiate $f$ , and we can integrate $g$ . Integration by parts exchanges the problem of computing $$\int fg dx\tag 1$$ with that of computing $$-\int f'G dx\tag 2$$ plus an extra term, where $G$ is an antiderivative of $g$ . This can be repeated to get to $$(-1)^n\int f^{(n)}g^{(-n)} dx\tag 3$$ again plus some other terms, where $f^{(n)}$ is the $n$ -fold derivative of $f$ , and $g^{(-n)}$ is the $n$ -fold antiderivative of $g$ . So, applying this rule $n$ -times is useful precisely when $(3)$ happens to be easier than $(1)$ . Some cases when this happens: If $f$ is a polynomial, then $f^{(n)}=0$ for some $n$ , and $(3)$ vanishes and is therefore easier than $(1)$ . I

Cites J.Doe · 2019-06 · Mathematics Stack Exchange · on “D I - Method or Tabulator method collapses at $\sin^2(x)$ - why is that?”

I have recently learned about the tabulator or D-I Method [this video] . This method - as far as I got it right - is just a variation of integration by parts. It works out very well, even in cases where you get a repetition in one line. But for $\int \sin^2(x) \ dx$ it doesn't work out and ends in $0=0$ . I tried to do it several times, checked each step and it doesn't work out. Why is that? Edit: Calculations (this is the same as when using the DI-Method): $$\int sin²(x) \\= -sin(x)*cos(x) - \int cos(x)*cos(x) \\=-sin(x)*cos(x)+cos(x)*sin(x) + \int cos(x)*sin(x) $$ at this spot you can see that each part is canceling out each other.

Cites Rounak Sarkar · 2021-06 · Mathematics Stack Exchange · on “On $\int_0^{\pi} \operatorname{Si}^n(x) \ \mathrm{d}x $”

I am not sure my result is entirely correct or not but if there is any mistakes then please tell me in the comments. So the integral is, $\textstyle\displaystyle{\int_{0}^{\pi}\operatorname{Si}^n(x)dx}$ I think we can start to simplify this using integration by parts, more specifically the DI method (If you don't know what DI method is then for introduction you can watch this [this video] and this [LINK] video. So let's start- $\begin{array} \ & \text{D} & \text{I}\\ + & \operatorname{Si}^n(x) & 1\\ - & n\operatorname{Si}^{n-1}(x)\frac{\sin(x)}{x} & x \end{array}$ $\textstyle\displaystyle{\therefore\int_{0}^{\pi}\operatorname{Si}^n(x)dx}$ $\textstyle\displaystyle{=x\operatorname{Si}^n(x)\bigg]_{0